High School

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29. Water rises in a capillary tube of radius r up to a height h. The mass of water in the capillary tube is m. The mass of water that will rise in a capillary tube of radius 2r will be:

(a) 2m
(b) m
(c) m/2
(d) 4m

Answer :

Let's explore the concept of capillary action and how it applies to this question.

Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of external forces like gravity. This occurs due to the adhesive forces between the liquid and the walls of the tube and the cohesive forces within the liquid.

For a liquid rising in a capillary tube, the height [tex]h[/tex] to which the liquid rises is given by the formula:
[tex]h = \frac{2T \cos \theta}{\rho g r}[/tex]
where:

  • [tex]T[/tex] is the surface tension of the liquid,
  • [tex]\theta[/tex] is the contact angle,
  • [tex]\rho[/tex] is the density of the liquid,
  • [tex]g[/tex] is the acceleration due to gravity,
  • [tex]r[/tex] is the radius of the tube.

The mass [tex]m[/tex] of the liquid in the tube is given by the volume of the liquid column times its density:
[tex]m = \pi r^2 h \rho[/tex]

Now, let's suppose we use a tube with radius [tex]2r[/tex]. Using the same formula for height [tex]h[/tex] (which still holds since the angle, surface tension, gravity, and density remain constant), we find that [tex]h[/tex] decreases as [tex]r[/tex] increases:
[tex]h' = \frac{2T \cos \theta}{\rho g (2r)} = \frac{h}{2}[/tex]

Therefore, the new height [tex]h'[/tex] will be half of the original height. The mass of water in the tube with this new height can be expressed as:
[tex]m' = \pi (2r)^2 h' \rho = 4\pi r^2 \frac{h}{2} \rho = 2 \times \pi r^2 h \rho[/tex]

The mass of water in the tube with radius [tex]2r[/tex] will be [tex]2m[/tex].

Thus, the correct answer is option (a) [tex]2m[/tex].

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Rewritten by : Barada